Answer: The final volume reading on the base is 72.35 mL
Explanation:
To calculate the volume of base, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]H_2SO_4[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is [tex]Ba(OH)_2[/tex]
We are given:
[tex]n_1=2\\M_1=0.472M\\V_1=[41.90-34.42]=7.48mL\\n_2=2\\M_2=0.388M\\V_2=?mL[/tex]
Putting values in above equation, we get:
[tex]2\times 0.472\times 7.48=2\times 0.388\times V_2\\\\V_2=\frac{2\times 0.472\times 7.48}{2\times 0.388}=9.10mL[/tex]
Final volume of barium hydroxide solution = Initial volume + Volume needed to neutralize = [63.25 + 9.10] = 72.35 mL
Hence, the final volume reading on the base is 72.35 mL